# 离散数学 Discrete Mathematics MATH223001/MATH223101

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We apply the lower bound in Lemma 2.5.1 to the logarithms. For a typical term, we get
$$\ln \left(\frac{m+t-k}{m-k}\right) \geq \frac{\frac{m+t-k}{m-k}-1}{\frac{m+t-k}{m-k}}=\frac{t}{m+t-k}$$

and so
\begin{aligned} \ln \left(\frac{m+t}{m}\right) &+\ln \left(\frac{m+t-1}{m-1}\right)+\cdots+\ln \left(\frac{m+1}{m-t+1}\right) \ & \geq \frac{t}{m+t}+\frac{t}{m+t-1}+\cdots+\frac{t}{m+1} . \end{aligned}
We replace each denominator by the largest one to decrease the sum:
$$\frac{t}{m+t}+\frac{t}{m+t-1}+\cdots+\frac{t}{m+1} \geq \frac{t^{2}}{m+t} .$$

## MATH223001/MATH223101COURSE NOTES ：

For $n=1$ and $n=2$, if we require that $L_{n}$ be of the given form, then we get
$$L_{1}=1=a+b, \quad L_{2}=3=a \frac{1+\sqrt{5}}{2}+b \frac{1-\sqrt{5}}{2}$$
Solving for $a$ and $b$, we get
$$a=\frac{1+\sqrt{5}}{2}, \quad b=\frac{1-\sqrt{5}}{2}$$
Then
$$L_{n}=\left(\frac{1+\sqrt{5}}{2}\right)^{n}+\left(\frac{1-\sqrt{5}}{2}\right)^{n}$$

# 离散数学 Discrete Mathematics MATH00103

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translates into
$$c \cdot q^{n+1}=c \cdot q^{n}+c \cdot q^{n-1}$$
which after simplification becomes
$$q^{2}=q+1$$

So both numbers $c$ and $n$ disappear. ${ }^{1}$
So we have a quadratic equation for $q$, which we can solve and get
$$q_{1}=\frac{1+\sqrt{5}}{2} \approx 1.618034, \quad q_{2}=\frac{1-\sqrt{5}}{2} \approx-0.618034 .$$
This gives us two kinds of geometric progressions that satisfy the same recurrence as the Fibonacci numbers:
$$G_{n}=c\left(\frac{1+\sqrt{5}}{2}\right)^{n}, \quad G_{n}^{\prime}=c\left(\frac{1-\sqrt{5}}{2}\right)^{n}$$

## MATH00103 COURSE NOTES ：

(a) if $a \mid b$ and $b \mid c$ then $a \mid c$;
(b) if $a \mid b$ and $a \mid c$ then $a \mid b+c$ and $a \mid b-c$;
(c) if $a, b>0$ and $a \mid b$ then $a \leq b$;
(d) if $a \mid b$ and $b \mid a$ then either $a=b$ or $a=-b$.

# 离散数学|Discrete Mathematics代写

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Unfortumately, the answer is yes. The smallest such number is $341=11 \cdot 31$. This is not a prime, but it satisfies
$$341 \mid 2^{340}-1 .$$
(How do we know that this divisibility relation holds without extensive computation? We can use Fermat’s Theorem. It is sufficient to argue that both 11 and 31 are divisors of $2^{340}-1$, since then so is their product, 11 and 31 being different primes. By Fermat’s Theorem,
$$11 \mid 2^{10}-1$$

Next we invoke the result : It implies that
$$2^{10}-1 \mid 2^{340}-1$$
Hence
$$11 \mid 2^{340}-1$$
For 31, we don’t need Fermat’s Theorem, but only again:
$$31=2^{5}-1 \mid 2^{340}-1$$

## Oxford COURSE NOTES ：

4.3.2. For $n=1$ and $n=2$, if we require that $L_{n}$ be of the given form, then we get
$$L_{1}=1=a+b, \quad L_{2}=3=a \frac{1+\sqrt{5}}{2}+b \frac{1-\sqrt{5}}{2}$$
Solving for $a$ and $b$, we get
$$a=\frac{1+\sqrt{5}}{2}, \quad b=\frac{1-\sqrt{5}}{2}$$
Then
$$L_{n}=\left(\frac{1+\sqrt{5}}{2}\right)^{n}+\left(\frac{1-\sqrt{5}}{2}\right)^{n}$$

# 离散数学|Discrete Mathematics代写 5CCM251A

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3.8.2. We apply the lower bound in Lemma 2.5.1 to the logarithms. For a typical term, we get
$$\ln \left(\frac{m+t-k}{m-k}\right) \geq \frac{\frac{m+t-k}{m-k}-1}{\frac{m+t-k}{m-k}}=\frac{t}{m+t-k}$$

and so
\begin{aligned} \ln \left(\frac{m+t}{m}\right) &+\ln \left(\frac{m+t-1}{m-1}\right)+\cdots+\ln \left(\frac{m+1}{m-t+1}\right) \ & \geq \frac{t}{m+t}+\frac{t}{m+t-1}+\cdots+\frac{t}{m+1} \end{aligned}

## 5CCM251ACOURSE NOTES ：

For $n=1$ and $n=2$, if we require that $L_{n}$ be of the given form, then we get
$$L_{1}=1=a+b, \quad L_{2}=3=a \frac{1+\sqrt{5}}{2}+b \frac{1-\sqrt{5}}{2}$$
Solving for $a$ and $b$, we get
$$a=\frac{1+\sqrt{5}}{2}, \quad b=\frac{1-\sqrt{5}}{2}$$
Then
$$L_{n}=\left(\frac{1+\sqrt{5}}{2}\right)^{n}+\left(\frac{1-\sqrt{5}}{2}\right)^{n}$$

# 离散数学代写|discrect mathematics代写CMTH110代考

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This course covers the fundamentals of discrete mathematics with a focus on proof methods. Topics include: propositional and predicate logic, notation for modern algebra, naive set theory, relations, functions and proof techniques.

The number of elements of the set $X$, written $|X|$, is its cardinality.
From the definition, $|{1,2,3}|=3$. Also, $|\varnothing|=0$.
Definition 1.2 1. Given sets $X$ and $Y$, we say $X$ is a subset of $Y$ if every element of $X$ is also an element of $Y$. We write $X \subseteq Y$, and refer to the symbol $\subseteq$ as inclusion.

1. If there is an element of $Y$ that is not an element of $X$, we say that $X$ is a proper subset of $Y$, written $X \nsubseteq Y$. Note that if $X \varsubsetneqq Y$, then we also have that $X \subseteq Y$.
Example $1.2$ The set
$${1,3,4} \subseteq{1,2,3,4}$$
and it is also the case that
$${1,3,4} \varsubsetneqq{1,2,3,4}$$

These are two equations in two unknowns, and we solve them below
Subtracting (1.4) from (1.3), we obtain
$$V_{1}(H)-V_{1}(T)=\Delta_{0}\left(S_{1}(H)-S_{1}(T)\right)$$
so that
$$We may put sets together using several operations. Definition 1.6 Let X and Y be sets in a universe U. 1. The intersection of X and Y, written X \cap Y, is the set consisting of elements in both X and Y. We write$$
X \cap Y={u \in U: u \in X \text { and } u \in Y}
$$2. The union of X and Y, written X \cup Y, is the set consisting of elements in X or Y. We write$$
X \cup Y={u \in U: u \in X \text { or } u \in Y}
$$3. The difference of X with Y, written X \backslash Y, is the set consisting of elements in X but not in Y. We write$$
X \backslash Y={u \in U: u \in X \text { and } u \notin Y}
$$4. The complement of X, written X^{c}, is the set of elements of U not in X. We write$$
X^{c}={u \in U: u \notin X}


## CMTH110 COURSE NOTES ：

Definition 1.1. A statement (or proposition) is a sentence that is either true or false but not both.
Examples of statements:

• $p=$ “Alice is 12 years old”,
• $q=$ “Bob ate pizza for dinner yesterday”,
• $r=” 3 \cdot 3=9$ “,
• $s=” 2 \cdot 4=9 “$.
Examples of non-statements:
• “Is the sky blue?” (question)
• ” $x^{2}+y^{2}=13 “$ (for some values of $x$ and $y$ the proposition is true, whereas for others it is false).
Compound Statements
We can make new statement from old ones using operators like “not”, “and”, “or”, “if then”. There are a few simple examples:
• “Alice is not 12 years old” $(\sim p)$,
• “Bob ate pizza for dinner yesterday and $3 \cdot 3=9$ ” $(q \wedge r)$.
Compound statements (since they are (regular) statements as well) must have well-defined truth values – they must be either true or false. We can write down a table of truth values for compound statements based on the truth values of component statements.
Definition 1.2. The negation of a statement $p$ (denoted $\sim p$ ) is true if $p$ is false. If $p$ is true, then $\sim p$ is false.